0.00/0.20	YES
0.00/0.20	
0.00/0.20	DP problem for innermost termination.
0.00/0.20	P =
0.00/0.20	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
0.00/0.20	  f6#(I0, I1, I2) -> f5#(rnd1, 1, I2) [y1 = -1 /\ rnd1 = 1]
0.00/0.20	  f2#(I3, I4, I5) -> f3#(I3, I4, 0) 
0.00/0.20	  f5#(I6, I7, I8) -> f3#(I6, I7, 1) [1 <= I6 /\ I6 <= 1]
0.00/0.20	  f5#(I9, I10, I11) -> f1#(I9, I10, I11) [1 + I9 <= 1]
0.00/0.20	  f5#(I12, I13, I14) -> f1#(I12, I13, I14) [2 <= I12]
0.00/0.20	  f1#(I18, I19, I20) -> f3#(I18, I19, 1) [2 <= I18 /\ I18 <= 2]
0.00/0.20	  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I21 <= 2]
0.00/0.20	  f1#(I24, I25, I26) -> f2#(I24, I25, I26) [3 <= I24]
0.00/0.20	R = 
0.00/0.20	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
0.00/0.20	  f6(I0, I1, I2) -> f5(rnd1, 1, I2) [y1 = -1 /\ rnd1 = 1]
0.00/0.20	  f2(I3, I4, I5) -> f3(I3, I4, 0) 
0.00/0.20	  f5(I6, I7, I8) -> f3(I6, I7, 1) [1 <= I6 /\ I6 <= 1]
0.00/0.20	  f5(I9, I10, I11) -> f1(I9, I10, I11) [1 + I9 <= 1]
0.00/0.20	  f5(I12, I13, I14) -> f1(I12, I13, I14) [2 <= I12]
0.00/0.20	  f3(I15, I16, I17) -> f4(I15, I16, I17) 
0.00/0.20	  f1(I18, I19, I20) -> f3(I18, I19, 1) [2 <= I18 /\ I18 <= 2]
0.00/0.20	  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I21 <= 2]
0.00/0.20	  f1(I24, I25, I26) -> f2(I24, I25, I26) [3 <= I24]
0.00/0.20	
0.00/0.20	The dependency graph for this problem is:
0.00/0.20	  0 -> 1
0.00/0.20	  1 -> 3
0.00/0.20	  2 -> 
0.00/0.20	  3 -> 
0.00/0.20	  4 -> 7
0.00/0.20	  5 -> 6, 8
0.00/0.20	  6 -> 
0.00/0.20	  7 -> 2
0.00/0.20	  8 -> 2
0.00/0.20	Where:
0.00/0.20	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
0.00/0.20	  1) f6#(I0, I1, I2) -> f5#(rnd1, 1, I2) [y1 = -1 /\ rnd1 = 1]
0.00/0.20	  2) f2#(I3, I4, I5) -> f3#(I3, I4, 0) 
0.00/0.20	  3) f5#(I6, I7, I8) -> f3#(I6, I7, 1) [1 <= I6 /\ I6 <= 1]
0.00/0.20	  4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) [1 + I9 <= 1]
0.00/0.20	  5) f5#(I12, I13, I14) -> f1#(I12, I13, I14) [2 <= I12]
0.00/0.20	  6) f1#(I18, I19, I20) -> f3#(I18, I19, 1) [2 <= I18 /\ I18 <= 2]
0.00/0.20	  7) f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I21 <= 2]
0.00/0.20	  8) f1#(I24, I25, I26) -> f2#(I24, I25, I26) [3 <= I24]
0.00/0.20	
0.00/0.20	We have the following SCCs.
0.00/0.20	  
0.00/3.18	EOF